Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 4 - Types of Chemical Reactions and Solution Stoichiometry - Exercises - Page 181: 39


4.5 M

Work Step by Step

Sodium carbonate: $Na_2CO_3\Rightarrow 2Na^++CO_3^{2-}$ Sodium bicarbonate: $NaHCO_3\Rightarrow Na^{+}+HCO_3^{-}$ Moles of sodium ion = $0.070 \text{ L}*3.0\text{ M}*2+0.030 \text{ L}*1.0\text{ M}*1=0.45$ moles Total liters of solution = 0.0700 L + 0.0300 L = 0.100 L Sodium ion concentration = $\frac{0.45 \text{ moles}}{0.100 \text{ L}}=4.5\text{ M}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.